question_answer

Show that the slope of an adiabatic curve at any point is y times the slope of an isothermal curve at the corresponding point.             

Answer:

                For an isothermal change, \[PV=K\] Differentiating both sides, we get \[P.dV+V.dP=0\] or \[V.dP=-PdV\] \[\therefore \] Slope of an isothermal curve,   \[{{\left( \frac{dP}{dV} \right)}_{iso}}=-\frac{P}{V}\] For an adiabatic change,  \[P{{V}^{\gamma }}=K'\] Differentiating both sides, we get \[P.\gamma {{V}^{\gamma -1}}.dV+{{V}^{\gamma }}.dP=0\] or \[\gamma PdV+VdP=0\] or \[VdP=-\gamma PdV\] \[\therefore \]  Slope of an adiabatic curve, \[{{\left( \frac{dP}{dV} \right)}_{adia}}=-\frac{\gamma P}{V}\] Clearly, slope of an adiabatic curve \[=\gamma \times \]slope of an isothermal curve. As \[\gamma >1,\]so an adiabatic \[P-V\] curve is steeper than the corresponding isothermal \[P-V\] curve.